Question

Find The area of a triangle whose vertices are (5,0),(8,0) and (8,4)​

Answer 1

Answer:

6 sq.units

Step-by-step explanation:

find the length of the sides using distance formula for the three vertices. you'll find the triangle to be a right angled triangle so ; u can take any side as the height another the base but make sure not to take the hypotenuse as the base or height

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Area\:of\:triangle=6\:sq\:units}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (5,0) } \\ \\ \tt{: \implies Coordinate \: of \: B = (8,0) } \\ \\ \tt{: \implies Coordinate \: of \: C = (8,4) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |5(0 - 4) + 8(4 - 0) + 8(0 - 0)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |5 \times -4 + 8\times 4 + 8 \times 0 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-20+32 + 0| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 12} \\ \\ \green{\tt{: \implies Area \: of \: triangle =6 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$